Intercepts on the Axes made by a Circle

We will learn how to find the intercepts on the axes made by a circle.

The lengths of intercepts made by the circle x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 with X and Y axes are 2\(\mathrm{\sqrt{g^{2} - c}}\) and 2\(\mathrm{\sqrt{f^{2} - c}}\) respectively.

Proof:

Let the given equation of the circle be x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 ………. (1)

Clearly, the centre of the circle is c (-g, -f) and the radius = \(\mathrm{\sqrt{g^{2} + f^{2}- c}}\)

Let AB be the intercept made by the given circle on x-axe. Since on x-axis, y = 0. Therefore, x-coordinates of the points A and B are the roots of the equation x\(^{2}\) + 2gx + c = 0.

Let x\(_{1}\) and x\(_{2}\) be the x-coordinates of the points A and B respectively. Then, x\(_{1}\) and x\(_{2}\) also the roots of the equation x\(^{2}\) + 2gx + c = 0.

Therefore, x\(_{1}\) + x\(_{2}\) = - 2g and x\(_{1}\)x\(_{2}\) = c

Clearly the intercept on x-axis = AB

                                          = x\(_{2}\) - x\(_{1}\) = \(\mathrm{\sqrt{(x_{2} - x_{1})^{2}}}\)

                                          = \(\mathrm{\sqrt{(x_{2} + x_{1})^{2} - 4x_{1}x_{2}}}\)

                                          = \(\mathrm{\sqrt{4g^{2} - 4c}}\)

                                          = 2\(\mathrm{\sqrt{g^{2} - c}}\)

Therefore, the intercept made by the circle (1) on the x-axis = 2\(\mathrm{\sqrt{g^{2} - c}}\)

Again,

Let DE be the intercept made by the given circle on y-axe. Since on y-axis, x = 0. Therefore, y-coordinates of the points D and E are the roots of the equation y\(^{2}\) + 2fy + c = 0.

Let y\(_{1}\) and y\(_{2}\) be the x-coordinates of the points D and E respectively. Then, y\(_{1}\) and y\(_{2}\) also the roots of the equation y\(^{2}\) + 2fy + c = 0

Therefore, y\(_{1}\) + y\(_{2}\) = - 2f and y\(_{1}\)y\(_{2}\) = c

Clearly the intercept on y-axis = DE

                                          = y\(_{2}\) - y\(_{1}\) = \(\mathrm{\sqrt{(y_{2} - y_{1})^{2}}}\)

                                          = \(\mathrm{\sqrt{(y_{2} + y_{1})^{2} – 4y_{1}y_{2}}}\)

                                          = \(\mathrm{\sqrt{4f^{2} - 4c}}\)

                                          = 2\(\mathrm{\sqrt{f^{2} - c}}\)

Therefore, the intercept made by the circle (1) on the y-axis = 2\(\mathrm{\sqrt{f^{2} - c}}\)

Solved examples to find the intercepts made by a given circle on the co-ordinate axes:

1. Find the length of the x-intercept and y-intercept made by the circle x\(^{2}\) + y\(^{2}\) - 4x -6y - 5 = 0 with the co-ordinate axes.

Solution:

Given equation of the circle is x\(^{2}\) + y\(^{2}\) - 4x -6y - 5 = 0.

Now comparing the given equation with the general equation of the circle x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0, we get g = -2 and f = -3 and c = -5

Therefore, length of the x-intercept = 2\(\mathrm{\sqrt{g^{2} - c}}\) = 2\(\mathrm{\sqrt{4 - (-5) }}\) = 2√9 = 6.

The length of the y-intercept = 2\(\mathrm{\sqrt{f^{2} - c}}\) = 2\(\mathrm{\sqrt{9 - (-5) }}\) = 2√14.


2. Find the equation of a circle which touches the y-axis at a distance -3 from the origin and cuts an intercept of 8 units with the positive direction of x-axis.

Solution:

Let the equation of the circle be x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 …………….. (i)

According to the problem, the equation (i) touches the y-axis

Therefore, c = f\(^{2}\) ………………… (ii)

Again, the point (0, -3) lies on the circle (i).

Therefore, putting the value of x = 0 and y = -3 in (i) we get,

9 - 6f + c = 0 …………………… (iii)

From (ii) and (iii), we get 9 - 6f + f\(^{2}\) = 0 ⇒ (f - 3)\(^{2}\) = 0 ⇒ f - 3 = 0 ⇒ f = 3

Now putting f = 3 in (i) we get, c = 9

Again, according to the problem the equation of the circle (i) cuts an intercept of 8 units with the positive direction of x-axis.

Therefore,

2\(\mathrm{\sqrt{g^{2} - c}}\) = 8

⇒ 2\(\mathrm{\sqrt{g^{2} - 9}}\) = 8

⇒ \(\mathrm{\sqrt{g^{2} - 9}}\) = 4

⇒ g\(^{2}\) - 9 = 16, [Squaring both sides]

⇒ g\(^{2}\) = 16 + 9

⇒ g\(^{2}\) = 25

⇒ g = ±5.

Hence, the required equation of the circle is x^2 + y^2 ± 10x + 6y + 9 = 0.

 The Circle




11 and 12 Grade Math 

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